Dr. Hackney STA Solutions pg 13

# Dr. Hackney STA Solutions pg 13 - 1-10Solutions Manual for...

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Unformatted text preview: 1-10Solutions Manual for Statistical Inferencec. If all of theAiare equal, all of the probabilities in the inclusion-exclusion identity are thesame. ThusP1=nP(A),P2=n2P(A),...,Pj=njP(A),and the sequence of upper bounds onP(∪iAi) =P(A) becomesP1=nP(A),P1-P2+P3=n-n2+n3P(A),...which eventually sum to one, so the last bound is exact. For the lower bounds we getP1-P2=n-n2P(A),P1-P2+P3-P4=n-n2+n3-n4P(A),...which start out negative, then become positive, with the last one equalingP(A) (see Schwa-ger 1984 for details).1.44P(at least 10 correct|guessing) =∑20k=10(20k)(14)k(34)n-k=.01386.1.45Xis finite. ThereforeBis the set of all subsets ofX. We must verify each of the three propertiesin Definition 1.2.4. (1) IfA∈ BthenPX(A) =P(∪xi∈A{sj∈S:X(sj) =xi})≥0 sincePis a probability function. (2)PX(X) =P(∪mi=1{sj∈S:X(sj) =xi}) =P(S) = 1. (3) IfA1,A2,...∈ Band pairwise disjoint thenPX(∪∞k=1Ak)=P(∞[k=1{∪xi∈Ak{sj∈S:X(sj) =xi}})=∞Xk=1P(∪xi∈Ak{sj∈S:X(sj) =xi}) =∞Xk=1...
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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